The Gibbs free energy change of a reaction at `27^(@)C` is -26 Kcal. and its entropy change is -60 Cals/K. `Delta H` for the reaction is :-

Options

A. `-44` K. Cals.
B. `-18` K. Cals.
C. 34 K. Cals.
D. `-24` K. Cals.

Answer

To find the value of ΔH (enthalpy change) for the reaction, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔG is the Gibbs free energy change,
- ΔH is the enthalpy change,
- T is the temperature in Kelvin,
- ΔS is the entropy change.

### Step-by-Step Solution:

1. **Convert the given temperature to Kelvin**:
The temperature is given as 27°C. To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, \text{K} \]

2. **Convert ΔG from kilocalories to calories**:
The Gibbs free energy change (ΔG) is given as -26 kcal. To convert this to calories:
\[ \Delta G = -26 \, \text{kcal} \times 1000 \, \text{cal/kcal} = -26000 \, \text{cal} \]

3. **Identify the given entropy change (ΔS)**:
The entropy change is given as:
\[ \Delta S = -60 \, \text{cal/K} \]

4. **Substitute the values into the Gibbs free energy equation**:
Now we can rearrange the Gibbs free energy equation to solve for ΔH:
\[ \Delta H = \Delta G + T \Delta S \]
Substituting the values we have:
\[ \Delta H = -26000 \, \text{cal} + (300 \, \text{K}) \times (-60 \, \text{cal/K}) \]

5. **Calculate TΔS**:
\[ T \Delta S = 300 \times -60 = -18000 \, \text{cal} \]

6. **Combine the values**:
Now substitute this back into the equation for ΔH:
\[ \Delta H = -26000 \, \text{cal} - 18000 \, \text{cal} \]
\[ \Delta H = -44000 \, \text{cal} \]

7. **Convert ΔH back to kilocalories**:
To convert calories back to kilocalories:
\[ \Delta H = -44000 \, \text{cal} \div 1000 = -44 \, \text{kcal} \]

### Final Answer:
\[ \Delta H = -44 \, \text{kcal} \]